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ASHA 777 [7]
3 years ago
15

light incident on a polarizer is then passed through a second polarizer. if the polarizer and the analyzer are perpendicular to

each other, what can you say about the light waves emerging from the second polarizer?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
Physics
2 answers:
Lerok [7]3 years ago
8 0
<span>Indeed, this is one of the odd results in physics. A system of two polarizing filters arranged as shown below trasmits no light.</span>
rewona [7]3 years ago
8 0

Answer:

The intensity of light emerging from the second polarizer is zero.

Explanation:

The angle between the axis of first polarizer and the second polarizer is 90°.

According to the law of Malus

I = I_{0}\times Cos^{2}\Theta

where, I_{0} be the intensity of incident light on first polarizer and I be the intensity of light emerging out from second polarizer.

The intensity of light emerging out from first polarizer is given by

I_{1}=\frac{I_{0}}{2}

the inetnsity of light emerging out from second polarizer is given by

I = I_{1}\times Cos^{2}90

I = 0

Thus, no light waves are emerging out from the second polarizer.

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A rock rolls down a hill. Which form of energy is this an example of? (2 points)
Bond [772]

Answer:

c.Mechanical

Explanation:

6 0
3 years ago
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

\frac{1}{2}mv^2=0.0052

\frac{1}{2}\times 0.00524\times v^2=0.0052

v = 1.40 m/sec

7 0
3 years ago
A rocket is fired at 100 m/s at an angle of 37, how many seconds is it in the air?
Sonbull [250]

Answer:

12.3 seconds

Explanation:

time in air = 2(v)/g

2(100sin(37))/9.8

=12.28 seconds

7 0
3 years ago
A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike rider is 91kg. What is the kinetic energy of
Anestetic [448]

a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

KE = \frac{1}{2}m* v^{2}

It is given:

v = 8.5 m/s

m = 91 kg

So:

KE= \frac{1}{2}*91*8.5^{2} =3,287.4J


b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h

In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J

m = 91 kg

g = 9.81 m/s² - gravitational acceleration

h = ? - height


Now, let's replace those:

3,287.4= 91 * 9.81 * h

⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m

3 0
3 years ago
how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?
SCORPION-xisa [38]

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = -PE_g = -2058\ kg.m/s^2

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= \frac{1}{2} m\times v^2 = m\times g \times h

Solving for v, we get

v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2

The person will be going at a speed of 7.67 m/s².

4 0
3 years ago
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